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# Chapter 3: Trigonometric Identities and Equations

Difficulty Level: At Grade Created by: CK-12

## Chapter Summary

Here are the identities studied in this chapter:

Quotient & Reciprocal Identities

tanθcscθ=sinθcosθcotθ=cosθsinθ=1sinθ secθ=1cosθ cotθ=1tanθ

Pythagorean Identities

sin2θ+cos2θ=11+cot2θ=csc2θtan2θ+1=sec2θ

Even & Odd Identities

sin(x)csc(x)=sinx=cscxcos(x)=cosxsec(x)=secxtan(x)=tanxcot(x)=cotx

Co-Function Identities

sin(π2θ)=cosθcsc(π2θ)=secθcos(π2θ)=sinθsec(π2θ)=cscθtan(π2θ)=cotθcot(π2θ)=tanθ

Sum and Difference Identities

cos(α+β)sin(α+β)tan(α+β)=cosαcosβsinαsinβ=sinαcosβ+cosαsinβ=tanα+tanβ1tanαtanβcos(αβ)=cosαcosβ+sinαsinβsin(αβ)=sinαcosβcosαsinβtan(αβ)=tanαtanβ1+tanαtanβ

Double Angle Identities

cos(2α)sin(2α)tan(2α)=cos2αsin2α=2cos2α1=12sin2α=2sinαcosβ=2tanα1tan2α

Half Angle Identities

cosα2=±1+cosα2sinα2=±1cosα2tanα2=1cosαsinα=sinα1+cosα

Product to Sum & Sum to Product Identities

sina+sinbsinasinbcosa+cosbcosacosb=2sina+b2cosab2=2sinab2cosa+b2=2cosa+b2cosab2=22sina+b2sinab2sinasinb=12[cos(ab)cos(a+b)]cosacosb=12[cos(ab)+cos(a+b)]sinacosb=12[sin(a+b)+sin(ab)]cosasinb=12[sin(a+b)sin(ab)]

Linear Combination Formula

\begin{align*}A \cos x + B \sin x = C \cos (x - D)\end{align*}, where \begin{align*}C = \sqrt{A^2 + B^2}, \cos D = \frac{A}{C}\end{align*} and \begin{align*}\sin D = \frac{B}{C}\end{align*}

## Review Questions

1. Find the sine, cosine, and tangent of an angle with terminal side on \begin{align*}(-8, 15)\end{align*}.
2. If \begin{align*}\sin a = \frac{\sqrt{5}}{3}\end{align*} and \begin{align*}\tan a < 0\end{align*}, find \begin{align*}\sec a\end{align*}.
3. Simplify: \begin{align*}\frac{\cos^4 x - \sin^4 x}{\cos^2 x - \sin^2 x}\end{align*}.
4. Verify the identity: \begin{align*}\frac{1 + \sin x}{\cos x \sin x} = \sec x (\csc x + 1)\end{align*}

For problems 5-8, find all the solutions in the interval \begin{align*}[0, 2\pi)\end{align*}.

1. \begin{align*}\sec \left (x + \frac{\pi}{2} \right ) + 2 = 0\end{align*}
2. \begin{align*}8 \sin \left (\frac{x}{2} \right ) - 8 = 0\end{align*}
3. \begin{align*}2 \sin^2 x + \sin 2x =0\end{align*}
4. \begin{align*}3 \tan^2 2x = 1\end{align*}
5. Solve the trigonometric equation \begin{align*}1 - \sin x = \sqrt{3} \sin x\end{align*} over the interval \begin{align*}[0, \pi]\end{align*}.
6. Solve the trigonometric equation \begin{align*}2 \cos 3x - 1 = 0\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.
7. Solve the trigonometric equation \begin{align*}2 \sec^2 x - \tan^4 x = 3\end{align*} for all real values of \begin{align*}x\end{align*}.

Find the exact value of:

1. \begin{align*}\cos 157.5^\circ\end{align*}
2. \begin{align*}\sin \frac{13 \pi}{12}\end{align*}
3. Write as a product: \begin{align*}4(\cos 5x + \cos 9x)\end{align*}
4. Simplify: \begin{align*}\cos(x - y) \cos y - \sin(x - y) \sin y\end{align*}
5. Simplify: \begin{align*}\sin \left (\frac{4 \pi}{3} - x \right ) + \cos \left (x + \frac{5 \pi}{6} \right )\end{align*}
6. Derive a formula for \begin{align*}\sin 6x\end{align*}.
7. If you solve \begin{align*}\cos 2x = 2 \cos^2x - 1\end{align*} for \begin{align*}\cos^2 x\end{align*}, you would get \begin{align*}\cos^2 x = \frac{1}{2} (\cos 2x + 1)\end{align*}. This new formula is used to reduce powers of cosine by substituting in the right part of the equation for \begin{align*}\cos^2 x\end{align*}. Try writing \begin{align*}\cos^4 x\end{align*} in terms of the first power of cosine.
8. If you solve \begin{align*}\cos 2x = 1 - 2 \sin^2 x\end{align*} for \begin{align*}\sin^2x\end{align*}, you would get \begin{align*}\sin^2 x = \frac{1}{2} (1 - \cos 2x)\end{align*}. Similar to the new formula above, this one is used to reduce powers of sine. Try writing \begin{align*}\sin^4x\end{align*} in terms of the first power of cosine.
9. Rewrite in terms of the first power of cosine:
1. \begin{align*}\sin^2x \cos^2 2x\end{align*}
2. \begin{align*}\tan^4 2x\end{align*}

1. If the terminal side is on \begin{align*}(-8,15)\end{align*}, then the hypotenuse of this triangle would be 17 (by the Pythagorean Theorem, \begin{align*}c = \sqrt{(-8)^2 + 15^2}\end{align*}). Therefore, \begin{align*}\sin x = \frac{15}{17}, \cos x = - \frac{8}{17}\end{align*}, and \begin{align*}\tan x = - \frac{15}{8}\end{align*}.
2. If \begin{align*}\sin a = \frac{\sqrt{5}}{3}\end{align*} and \begin{align*}\tan a < 0\end{align*}, then \begin{align*}a\end{align*} is in Quadrant II. Therefore \begin{align*}\sec a\end{align*} is negative. To find the third side, we need to do the Pythagorean Theorem.
So \begin{align*}\sec a = -\frac{3}{2}\end{align*}.
3. Factor top, cancel like terms, and use the Pythagorean Theorem Identity. Note that this simplification doesn't hold true for values of \begin{align*}x\end{align*} that are \begin{align*}\frac{\pi}{4} + \frac{n\pi}{2}\end{align*}, where \begin{align*}n\end{align*} is a positive integer,, since the original expression is undefined for these values of \begin{align*}x\end{align*}.
4. Change secant and cosecant into terms of sine and cosine, then find a common denominator.
5. \begin{align*}\sin^{-1} \left (\frac{1}{1 + \sqrt{3}} \right ) = x\end{align*} or \begin{align*}x = .3747\end{align*} radians and \begin{align*}x = 2.7669\end{align*} radians
6. Because this is \begin{align*}\cos 3x\end{align*}, you will need to divide by 3 at the very end to get the final answer. This is why we went beyond the limit of \begin{align*}2\pi\end{align*} when finding \begin{align*}3x\end{align*}.
7. Rewrite the equation in terms of tan by using the Pythagorean identity, \begin{align*}1 + \tan^2 \theta = \sec^2 \theta\end{align*}.
Because these factors are the same, we only need to solve one for \begin{align*}x\end{align*}.
Where \begin{align*}k\end{align*} is any integer.
8. Use the half angle formula with \begin{align*}315^\circ\end{align*}.
9. Use the sine sum formula.
10. Use the sine and cosine sum formulas.
11. Use the sine sum formula as well as the double angle formula.
12. Using our new formula, \begin{align*}\cos^4 x = \left [\frac{1}{2}(\cos 2x + 1) \right ]^2 \end{align*} Now, our final answer needs to be in the first power of cosine, so we need to find a formula for \begin{align*}\cos^2 2x\end{align*}. For this, we substitute \begin{align*}2x\end{align*} everywhere there is an \begin{align*}x\end{align*} and the formula translates to \begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}. Now we can write \begin{align*}\cos^4 x\end{align*} in terms of the first power of cosine as follows.
13. Using our new formula, \begin{align*}\sin^4 x = \left [\frac{1}{2}(1 - \cos 2x) \right ]^2 \end{align*} Now, our final answer needs to be in the first power of cosine, so we need to find a formula for \begin{align*}\cos^2 2x\end{align*}. For this, we substitute \begin{align*}2x\end{align*} everywhere there is an \begin{align*}x\end{align*} and the formula translates to \begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}. Now we can write \begin{align*}\sin^4 x\end{align*} in terms of the first power of cosine as follows.
14. (a) First, we use both of our new formulas, then simplify:
(b) For tangent, we use the identity \begin{align*}\tan x = \frac{\sin x}{\cos x}\end{align*} and then substitute in our new formulas. \begin{align*}\tan^4 2x = \frac{\sin^4 2x}{\cos^4 2x} \rightarrow\end{align*} Now, use the formulas we derived in #18 and #19.

## Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9701.

Feb 23, 2012